# If $a^b\equiv a^c\equiv 1\pmod{m}$ and $g=\gcd(b,c)$ prove $a^g\equiv 1\pmod{m}$

Let $$b,c,$$ and $$m$$ be positive integers and suppose $$a$$ is relatively prime to $$m$$. Furthermore, assume $$a^b\equiv a^c\equiv 1\pmod{m}$$ and that $$g=\gcd(b,c)$$.

I know that since $$g=\gcd(b,c)$$ it follows that $$g|b$$ and $$g|c$$. So one could write $$b=gx$$ and $$c=gy$$ for some $$x,\;y\;\epsilon\;\mathbb Z$$. I then have $$a^b\equiv a^{gx}\equiv (a^{g})^x\equiv 1\pmod{m}$$. I don't quite understand how to prove $$a^g\equiv 1\pmod{m}$$. I'm wondering if I should be utilizing the fact that the order of $$a$$ divides any power, $$k$$, such that $$a^k\equiv 1\pmod{m}$$. Or if I was on the right track just need to manipulate $$b$$ and $$c$$ slightly differently. Any help would be much appreciated.

Hint $$\,\ a^{\large b}\equiv 1\equiv a^{\large c}\iff {\rm ord}\,a\mid b,c\iff {\rm ord}\,a\mid \gcd(b,c)\iff a^{\large \gcd(b,c)}\equiv 1$$
• See here for the ${\rm ord}$ law used above, and see here for the $\gcd$ law used. We could instead use the Bezout identity for the gcd, but that is less conceptual (and less general). Feb 23, 2019 at 22:59
• Can I use the fact that g=bx+cy to rewrite this as $a^g \equiv a^{bx+cy}\equiv a^{bx}a^{cy}\equiv (a^b)^x(a^c)^y\equiv 1^x*1^y \equiv 1 \pmod{m}$ Feb 23, 2019 at 23:10
• @mjoseph Yes, that's the Bezout-based method that I alluded to above. Note that some exponents are negative so you need to remark that $a$ is invertible because $\ldots$ Feb 23, 2019 at 23:14